# How do I myself compute the following probability I tried to approximate binomial to a poisson distribution but this did not match the conditions since n is too small. Afterwards I considered the event Y equal the square root of X as being an intersection of 2 independents events Which means this is equal to the probability of y times the probability of square root of x. however, here the square root really confuses me and in general I have thought about it too much but I am stuck.

• Before posting another question it would be nice checking the answers/giving a reply to the answers of your previous questions. – callculus Jul 14 '18 at 16:10
• I am so sorry if I did not reply to the previous questions, but I can tell you that I checked all of the explanations. i can just say that this site is perfect. The people answering are experts and have deep knowledge of probability. I am really stunned. I’ve been stuck for many day on those questions and they help me solve it in just few minutes – Roy Rizk Jul 14 '18 at 16:15
• You can mark the questions as answered by clicking on the checkmark at your favorable answer. The advantage for all users is, that everybody can see which questions are answered and which questions are not answered. – callculus Jul 14 '18 at 16:20
• In R: i = c(0,1,4); sum(dbinom(i,7,1/4)*dpois(sqrt(i), 3)) returns 0.0660885. Same as @BrianTung's Answer (+1). Also less elegantly, an approximation by simulation mean(replicate(10^6, rbinom(1, 7, 1/4)==(rpois(1,3))^2)) returns 0.066067, correct to three places. – BruceET Jul 14 '18 at 16:38

## 2 Answers

\begin{eqnarray*} P(\sqrt{X}=Y) &=& \sum _{k=0}^2 P(\{\sqrt{X}=k\}\cap \{Y=k\})\\ &=& \sum _{k=0}^2 P(\sqrt{X}=k)P(Y=k)\\ &=& \sum _{k=0}^2 P(X=k^2)P(Y=k)\\ \end{eqnarray*}

Note that $X$ is constrained to be an integer between $0$ and $7$, and $Y$ is constrained to be an integer. That means that for $\sqrt{X} = Y$ to hold, $X$ must be $0$, $1$, or $4$. I would compute these probabilities explicitly and just calculate

$$P(X = 0)P(Y = 0) + P(X = 1)P(Y = 1) + P(X = 4)P(Y = 2)$$